find-the-domain-and-range-of-y-4x-2-3-x-lt-2-3x-1-x-2-

Question Number 5237 by sanusihammed last updated on 02/May/16

find the domain and range of y = [_( 4x^2 +3 x <2) ^( 3x−1 x ≥ 2)

$${find}\:{the}\:{domain}\:{and}\:{range}\:{of}\: \\ $$$${y}\:=\:\left[_{\:\mathrm{4}{x}^{\mathrm{2}} +\mathrm{3}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}\:<\mathrm{2}} ^{\:\mathrm{3}{x}−\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}\:\geqslant\:\mathrm{2}} \right. \\ $$

Answered by FilupSmith last updated on 02/May/16

Domain = {x: −∞≤x≤∞} Range_1 y=3x−1 {y: [3(2)−1]≤y≤∞} {y: 5≤y≤∞} Range_2 y=4x^2 +3 {y: −∞≤y<[4(2)^2 +3]} {y: −∞≤y<19} ∴Range = {y: −∞≤y≤∞}

$${Domain}\:=\:\left\{{x}:\:−\infty\leqslant{x}\leqslant\infty\right\} \\ $$$$ \\ $$$${Range}_{\mathrm{1}} \:{y}=\mathrm{3}{x}−\mathrm{1} \\ $$$$\left\{{y}:\:\left[\mathrm{3}\left(\mathrm{2}\right)−\mathrm{1}\right]\leqslant{y}\leqslant\infty\right\} \\ $$$$\left\{{y}:\:\mathrm{5}\leqslant{y}\leqslant\infty\right\} \\ $$$$ \\ $$$${Range}_{\mathrm{2}} \:{y}=\mathrm{4}{x}^{\mathrm{2}} +\mathrm{3} \\ $$$$\left\{{y}:\:−\infty\leqslant{y}<\left[\mathrm{4}\left(\mathrm{2}\right)^{\mathrm{2}} +\mathrm{3}\right]\right\} \\ $$$$\left\{{y}:\:−\infty\leqslant{y}<\mathrm{19}\right\} \\ $$$$ \\ $$$$\therefore{Range}\:=\:\left\{{y}:\:−\infty\leqslant{y}\leqslant\infty\right\} \\ $$

Answered by Rasheed Soomro last updated on 02/May/16

y = [_( 4x^2 +3 x <2) ^( 3x−1 x ≥ 2) The function is defined for all real numbers. Hence domain of function is R. ^• For x≥2 x≥2 ⇒3x≥6 ⇒3x−1≥5 ⇒ y≥5 ^(• ) For x<2 x<2 ⇒x^2 <4 ⇒4x^2 <16 ⇒4x^2 +3<19 ⇒ y<19 So the range is {y ∣ y∈ R ∧ 5≤y<19 }

$${y}\:=\:\left[_{\:\mathrm{4}{x}^{\mathrm{2}} +\mathrm{3}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}\:<\mathrm{2}} ^{\:\mathrm{3}{x}−\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}\:\geqslant\:\mathrm{2}} \right. \\ $$$$\mathrm{The}\:\mathrm{function}\:\mathrm{is}\:\mathrm{defined}\:\mathrm{for}\:\mathrm{all}\:\mathrm{real} \\ $$$$\mathrm{numbers}.\:\mathrm{Hence}\:\mathrm{domain}\:\mathrm{of}\:\mathrm{function}\: \\ $$$$\mathrm{is}\:\mathbb{R}. \\ $$$$\:^{\bullet} \mathrm{For}\:\mathrm{x}\geqslant\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{x}\geqslant\mathrm{2} \\ $$$$\:\:\:\:\:\Rightarrow\mathrm{3x}\geqslant\mathrm{6} \\ $$$$\:\:\:\:\:\Rightarrow\mathrm{3x}−\mathrm{1}\geqslant\mathrm{5} \\ $$$$\:\:\:\:\:\Rightarrow\:\mathrm{y}\geqslant\mathrm{5} \\ $$$$\:^{\bullet\:} \mathrm{For}\:\mathrm{x}<\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{x}<\mathrm{2} \\ $$$$\:\:\:\:\:\Rightarrow\mathrm{x}^{\mathrm{2}} <\mathrm{4} \\ $$$$\:\:\:\:\:\Rightarrow\mathrm{4x}^{\mathrm{2}} <\mathrm{16} \\ $$$$\:\:\:\:\:\Rightarrow\mathrm{4x}^{\mathrm{2}} +\mathrm{3}<\mathrm{19} \\ $$$$\:\:\:\:\:\Rightarrow\:\mathrm{y}<\mathrm{19} \\ $$$$\mathrm{So}\:\mathrm{the}\:\mathrm{range}\:\mathrm{is}\:\left\{\mathrm{y}\:\mid\:\mathrm{y}\in\:\mathbb{R}\:\wedge\:\:\mathrm{5}\leqslant\mathrm{y}<\mathrm{19}\:\right\} \\ $$

Commented by Yozzii last updated on 02/May/16

y≥3. Do a sketch graph for y to see this.

$${y}\geqslant\mathrm{3}.\:{Do}\:{a}\:{sketch}\:{graph}\:{for}\:{y}\:{to}\:{see} \\ $$$${this}. \\ $$

Commented by FilupSmith last updated on 02/May/16

$${Ah}\:{i}\:{miswrote}\:{the}\:{final}\:{part}! \\ $$

Commented by Rasheed Soomro last updated on 06/May/16

From graph I understand now that the range is {y: y∈R,y≥3} Credit goes to Yozzi. I think that for determining range of a function its minima and maxima should be considered.

$$\mathrm{From}\:\mathrm{graph}\:\mathrm{I}\:\mathrm{understand}\:\mathrm{now}\:\mathrm{that}\:\mathrm{the}\:\mathrm{range}\:\mathrm{is}\:\left\{{y}:\:{y}\in\mathbb{R},{y}\geqslant\mathrm{3}\right\} \\ $$$$\mathrm{Credit}\:\mathrm{goes}\:\mathrm{to}\:\mathrm{Yozzi}. \\ $$$$\mathrm{I}\:\mathrm{think}\:\mathrm{that}\:\mathrm{for}\:\mathrm{determining}\:\mathrm{range}\:\mathrm{of}\: \\ $$$$\mathrm{a}\:\mathrm{function}\:\mathrm{its}\:\mathrm{minima}\:\mathrm{and}\:\mathrm{maxima}\:\mathrm{should} \\ $$$$\mathrm{be}\:\mathrm{considered}. \\ $$

Commented by Rasheed Soomro last updated on 05/May/16

Commented by Yozzii last updated on 06/May/16

$${You}\:{also}\:{need}\:{to}\:{then}\:{find}\:{out}\:{whether} \\ $$$${those}\:{stationary}\:{points}\:{are}\:{global} \\ $$$${or}\:{local}\:{for}\:{all}\:{real}\:{x}.\:{If}\:{these}\:{points} \\ $$$${are}\:{global}\:{then}\:{you}\:{would}\:{have}\:{found} \\ $$$${one}\:{end}\:{of}\:{the}\:{interval}\:{for}\:{y}. \\ $$

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